FE Chemical Practice Question Solution included after investigation
CASE E-063 · FE

The Heater That Needed 418 kW, Not 209

Heating the liquid required 209 kJ/kg. At 2 kg/s, that became 418 kW of heat transfer.

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Case E-063 engineering failure visual
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Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

A liquid stream flows at 2.0 kg/s and is heated from 20°C to 70°C. Take cp=4.18 kJ/(kg·K), neglect phase change, kinetic and potential energy changes, and heat loss. A calculation finds cpΔT=209 kJ/kg and labels the heater duty 209 kW. Determine the correct heat-transfer rate.

Material Energy Balances Chemical Design Verification Root Cause Analysis Safe Correction
E-063
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CASE CLOSED · ENGINEERING VERDICT

What actually failed — and what should change.

What failed

Specific sensible energy was reported as equipment power.

What was missed

Mass flow rate.

Corrected model

Use Qdot=m_dot cpΔT for the stated steady sensible-heating case.

QUICK ANSWER

The sensible-energy increase is cpΔT=4.18×50=209 kJ/kg. At 2 kg/s, the required heat-transfer rate is 2×209=418 kJ/s, or 418 kW.

ENGINEERING TAKEAWAY

Specific enthalpy or sensible-energy change must be multiplied by mass flow rate to obtain a steady energy rate.

CASE QUESTIONS

Questions this case should settle.

How do you calculate steady heater duty for sensible heating?

Use Qdot=m_dot cp(Tout-Tin) when cp is treated as constant and no phase change occurs.

Why is 209 kJ/kg not 209 kW?

kJ/kg is energy per unit mass; kW is energy per unit time.

NEXT FE PRACTICE QUESTION

The Evaporator That Removed 750 kg/h of Water

Material Energy Balances · 5–10 min

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